TITLE: Fields having exactly one quadratic extension (up to isomorphism)
QUESTION [7 upvotes]: Let $F$ be an infinite field such that $F$ has, up to field isomorphisms$^{[1]}$, exactly one extension $K/F$ of degree $2$. Does it imply  that $[\overline F : F]<\infty$ ? What happens if $F$ has characteristic $0$?

I don't think that this holds, even if characteristic $0$, but I didn't have an example of a field $F$ of characteristic $0$ ($\implies F$ is infinite) such that $[\overline F : F] = \infty$ and in $F^*$, any product of two non-squares is a square, and there is at least one non-square (see 4) below).

My thoughts:
1) The only example of such $F$ I know is $\Bbb R$: any quadratic extension embeds in $\Bbb C$, which has already degree $2$ over $\Bbb R$. My question is to know if other examples exist: if $[\overline F : F]<\infty$ then (by Artin-Schreier theorem) $F$ is a real closed field.
2) Any finite field $\Bbb F_q$ has exactly one extension of degree $n$ (namely $\Bbb F_{q^n}$), up to field isomorphisms, for every $n \geq 1$. 
3) It implies that all the quadratic extensions are isomorphic as fields, but this is not sufficient, precisely when $F$ has no quadratic extension, e.g. $\overline F=F$ (or $\bigcup_{n \geq 0} K_n$, with $K_0=\Bbb Q,K_{n+1}=\{x \in \Bbb C \mid x^2 \in K_n\}$).
4) In characteristic different from $2$, any quadratic extension of $F$ is separable and has the form $F(\sqrt a)$ where $\sqrt a \not \in F$.
Therefore, as mentioned here, the quadratic extensions of $F$ correpond to $A:=F^* / F^{*,2}$ where $F^{*,2} = \{x^2 \mid \in F^*\}$. 
Notice that $A$ is a $\Bbb F_2$-vector space, via $[a]_2 \cdot [x]_{F^{*,2}} = [x^a]_{F^{*,2}}$.
So the most interesting case is when we are looking for fields of characteristic $\neq 2$ such that $A=F^* / F^{*,2}$ has order $2$.
Equivalently, in $F^*$, any product of two non-squares is a square, and there is at least one non-square
(because $x,y$ non squares $\implies x,1/y$ non-squares $\implies x/y = a^2 \in F^{*,2} \implies [x]_{F^{*,2}} = [y]_{F^{*,2}}$).
Let $a$ be a non-square in $F^*$ and let $i = \sqrt a$. Showing that $F(i)$ is algebraically closed is not reasonable. Notice that this condition about $[F^* : F^{*,2}]=2$ is involved in 3. here, which is precisely the situation where $a=-1$ is not a square.
Then I thought to some extension $F$ of $K=\mathrm{Frac}(\Bbb R[x,y]/(x^2+y^2+1))$ since $-1$ is not a square in $K^*$ but is a sum of squares ; we just need $[F^* : F^{*,2}]=2$, and then it's a counterexample, since $F$ won't be a formally real field. I only know how to do $[F^* : F^{*,2}]=1$, see my $K_n$'s in 3).
5) I tried $F = \overline{\Bbb F_2}(t)$, because $t^{1/n}$ has degree $n$ for any $n$, so $[\overline F :F]=\infty$. I think that any extension $F\left(\sqrt{P(t)/Q(t)}\right)$ is isomorphic to $F(\sqrt t) = \overline{\Bbb F_2}(\sqrt t)$ when $P,Q \in \overline{\Bbb F_2}[t]$ i.e. $P/Q \in F$, because we have $\sqrt{a+b}=\sqrt a + \sqrt b$ in the sense that $x^2=a,y^2=b \implies (x+y)^2 = a+b$, so that $\sqrt{P(t)/Q(t)}$ is just a rational fraction in $\sqrt t$, i.e. belongs to $F(\sqrt t)$.
However, in characteristic $2$, it is not clear that all quadratic extensions arise as $F(\sqrt a)$ for some non-square $a \in F$.
6) In characteristic 0 (at least $\neq 2$), it is not clear that $F(\sqrt{t+1}) \not \cong F(\sqrt t)$. If there was a field isomorphism, then there is $u \in  F(\sqrt t)$ such that $u^2=t+1$, hence there are $a,b \in F[t]$ such that $(a(\sqrt t)/b(\sqrt t))^2 = t+1$, which yields
$a(x)^2=(x^2+1)b(x)^2$ as polynomials in $F[x]$...

$^{[1]}$ I'm only interested in fields isomorphisms, not in "field extensions" isomorphisms (i.e. not in $F$-algebras isomorphisms – these are equivalent to saying that $f : K \stackrel{\cong}{\to} K'$ commutes with the embeddings $i : F \to K$ and $i' : F \to K'$).

REPLY [5 votes]: Finite fields still give loads of examples. Let $\kappa$ be any finite field, and let $q$ be any prime (not necessarily distinct from the characteristic). Now take the union $\mathcal K$  of all fields $K_n$ for which $[K_n:\kappa]=q^n$. Then $\mathcal K$ is infinite, and will have a unique extension of every degree prime to $q$, in particular only one quadratic extension, if $q\ne2$.
EDIT — Addition:
On looking more closely at your question, I see that you make some guesses about characteristic two. There the quadratic-extension picture is simultaneously simpler and more complicated. When you adjoin the square root of something, you’re making an inseparable extension of degree $p$, and the truth of the matter is that when the base is perfect, as $\overline{\Bbb F_2}$ is, and your field is finitely generated over the base, and the transcendence degree is only one, then there is precisely one inseparable extension of degree $p$, indeed only one purely inseparable extension of degree $p^m$ for each $m$. In other words, starting with $k=\overline{\Bbb F_2}(t)$, no matter what rational function you adjoin the square root of, you get the same extension, $k^{1/2}$
On the other hand, there are infinitely many nonisomorphic quadratic separable extensions of $k=\overline{\Bbb F_2}(t)$, for instance the ones gotten by adjoining the roots of $X^2+t^mX+t$.